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Question #fdd32

1 Answer

Sep 5, 2017

$430 . g$ $430.g$

Explanation:

This is simple unit conversion!

Start by converting the density to something favorable:

$\frac{1.11 g}{c m^{3}} \cdot \frac{c m^{3}}{m L} = \frac{1.11 g}{m L}$ $(1.11g)/(cm^3)*(cm^3)/(mL) = (1.11g)/(mL)$

Now things don't seem confusing, do they?

$387 m L \cdot \frac{1.11 g}{m L} \approx 430 . g$ $387mL * (1.11g)/(mL) approx 430.g$

The period indicates three significant figures, what is required in this problem.

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